Question

$\int \frac{e^x}{(2+e^x)(e^x+1)} dx =$

(where C is a constant of integration.)

• A. $log(\frac{e^x+2}{e^x+1})+C$
• B. $log(\frac{e^x}{e^x+2})+C$
• C. $\frac{e^x+1}{e^x+2}+C$
• D. $log(\frac{e^x+1}{e^x+2})+C$

Answer 1

Answer: (D)

$\ln\left(\frac{e^x+1}{e^x+2}\right) + C$

Answer 2

Let the given integral be $I$. $I = \int \frac{e^x}{(2+e^x)(e^x+1)} dx$ We use the substitution method. Let $u = e^x$. Then, the differential $du$ is given by $du = \frac{d}{dx}(e^x) dx = e^x dx$.

Substitute $u$ and $du$ into the integral: $I = \int \frac{du}{(2+u)(u+1)}$ Now, we have a rational function in terms of $u$. We can solve this integral using partial fraction decomposition. We decompose the integrand $\frac{1}{(u+2)(u+1)}$ into partial fractions: $\frac{1}{(u+2)(u+1)} = \frac{A}{u+2} + \frac{B}{u+1}$ To find the constants $A$ and $B$, we multiply both sides by $(u+2)(u+1)$: $1 = A(u+1) + B(u+2)$ Set $u = -1$: $1 = A(-1+1) + B(-1+2)$ $1 = A(0) + B(1)$ $1 = B$ Set $u = -2$: $1 = A(-2+1) + B(-2+2)$ $1 = A(-1) + B(0)$ $1 = -A$ $A = -1$ So, the partial fraction decomposition is: $\frac{1}{(u+2)(u+1)} = \frac{-1}{u+2} + \frac{1}{u+1}$ Now, we can rewrite the integral as: $I = \int \left(\frac{1}{u+1} - \frac{1}{u+2}\right) du$ Integrate each term: $I = \int \frac{1}{u+1} du - \int \frac{1}{u+2} du$ $I = \ln|u+1| - \ln|u+2| + C$ Using the property of logarithms $\ln a - \ln b = \ln \frac{a}{b}$: $I = \ln\left|\frac{u+1}{u+2}\right| + C$ Finally, substitute back $u = e^x$: $I = \ln\left|\frac{e^x+1}{e^x+2}\right| + C$ Since $e^x > 0$ for all real $x$, both $e^x+1$ and $e^x+2$ are positive. Thus, the absolute value is not necessary. $I = \ln\left(\frac{e^x+1}{e^x+2}\right) + C$

The final answer is $\ln\left(\frac{e^x+1}{e^x+2}\right) + C$.